I am still struggling quite a lot with the Picard-Lindelöf-Theorem (also known as the Cauchy-Lipschitz-Theorem).
Problem: Consider the following IVP with $\alpha \neq 1$ $$\begin{cases} y'&= t|y|^\alpha \\ y(0)&=1 \end{cases} $$ And show that there exists a unique solution $f_\alpha$ on an Interval $I_\alpha$ with $0 \in I_\alpha$, also show that $f_\alpha (t) > 0$ for small $t$
So to use Picard-Lindelöf I want to show that $f(t,y)=t|y|^\alpha$ is Lipschitz-continuous with respect to its second variable. I could do this by computing the partial derivative with respect to $y$ and find an interval on which this expression is bound, then on said hypothetical interval the solution would exist and be unique.
My approach: $$\frac{\partial f}{\partial y}= t\alpha|y|^{\alpha-1} \cdot \frac{y}{\sqrt{y^2}} $$ At which point I already run into trouble. I have made use of the fact that $|x|'= x/\sqrt{x^2}$ because it gives correct results in regard of the direction (when approaching from the left and the right). But the above expression is not defined at $y=0$.
So I thought if I could come up with an argument that shows why $y \neq 0$, then at least this problem would be out of the way.
Assume that $y=0$ then $y'=0=f(t,y) \implies \frac{\partial f}{\partial y}=0$ which is bound on $\mathbb{R} \implies f$ is Lipschitz continuous with respect to $y$ but we have $y(0)=1$ and thus $$y'=0 \implies y=c $$ So $y$ is a constant function and $y(0)=1$ means that $c=1 \implies y=1 \neq 0$ which shows that $y$ can not be the 'zero' function (if that exists).
Questions:
- Is the above argumentation correct?
- How can I find the desired interval $I_\alpha$ with $0 \in I_\alpha$?
Update: I tried to continue a bit on my own and ignore the Picard-Lindelöf part and just focus on solving the IVP, apparently the step I have taken above was necessary so I can divide by $|y|^\alpha$, however I will land at $$\frac{dy}{|y|^\alpha}=tdx $$ So it seems like I have to come up with a condition that guarantees me that $y$ is positive in some Intervall.